RONAN O'GORMAN
1. Introduction
Here we present the original proof of the following theorem by Erd®s, Rényi and Sós:
Theorem 1.1 (Friendship theorem). Suppose that, in a (nite) group of people, any two people have exactly one common friend. Then there is at least one "politician"
who is friends with everybody.
It is easy to see how one can interpret this in terms of graphs - let each of the people be represented by a vertex and draw an undirected edge between two vertices when the corresponding people are friends (assume no-one is friends with themselves). We say two vertices are neighbors" or adjacent" if there is an edge between them. The result can then be restated as follows:
Theorem 1.2 (Friendship theorem - reformulated). Let Gbe a graph with n ver- tices such that any two distinct vertices have exactly one common neighbor (we will refer to this as the friendship condition"). Then there is at least one vertex (a politician vertex") which is adjacent to every other.
It is easy to construct a graph of this form - take, for example, the windmill graph" shown in gure 1: In fact, we can show that any graph which satises our
Figure 1. Example of a windmill graph
friendship condition must have this form - and thus must have an odd number of vertices.
Remark 1.3. The qualiers nite" and exactly one" are important here - if we allow innite graphs, then starting from a ve-cycle, we can construct a counterexample by repeatedly add common neighbors for every pair of vertices that do not yet have
1
one. This leads to a countably innite graph satisfying the friendship condition with no politician.
The statement also fails if we allow more than one common neighbor - consider the four-cycle shown in gure 2
Figure 2. A 4-cycle
We follow Cathal Ó'Cléirigh's presentation of the proof, which is based on that of [1].
2. Proof of the friendship theorem
2.1. Outline. Our proof is a proof by contradiction. We will assume our graphG has no politician vertex and arrive at a contradiction by computing the degree" of each of its vertices:
Denition 2.1 (Degree of a vertex). The degree of a vertexuin Gis the number of vertices adjacent to it.
We proceed as follows:
(1) We show that any two non-adjacent vertices have equal degree.
(2) We extend this to show that all vertices have equal degreekfor somek∈N.
(3) We derive a formula for n (the number of vertices in the graph) in terms ofk, and show thatk >2.
(4) We compute eigenvalues of the adjacency matrix of G
(5) We consider the trace of the adjacency matrix and apply a theorem of Dedekind to obtain a contradiction.
2.2. Dedekind's theorem. For our last step, we shall need the following result from number theory, which is interesting in its own right:
Theorem 2.2 (Dedekind, 1858). Letm∈N. Then √
m∈Q =⇒ √ m∈N.
Proof. Assume that √
m ∈ Q. Let n0 be the smallest natural number with the property thatn0
√m∈N.
If √
m 6∈ N, then there exists some l ∈ N with 0 < √
m−l < 1. Let n = n0(√
m−l). Thenn < n0, and n√
m=n0
√m−n0l∈N,
a contradiction.
Step 1 - Non adjacent vertices have equal degree. Take any two non-adjacent verticesu, v in our graphG. Assumeuhas neighbors{w1,· · ·, wk}(sodegu=k).
Nowuandvhave exactly one common neighbor; re-labeling thewiif necessary we can assume that this isw2. Thenw2 andumust have a unique common neighbor, sow2is adjacent to exactly one of the otherwi; assume this isw1(again, re-labeling vertices if necessary).
Thenv has common neighborw2 withw1. It must also have a unique common neighbor zi with eachwi fori≥2. Ifzi =zj for some i6=j thenzj would have 2 common neighborswi andwj withu, so all thezi are distinct. Thereforev has at leastkneighbors {w2, z2, z3,· · ·, zk}so hencedegv≥degu(see gure 3).
Figure 3.
Similarly, we can deduce thatdegu≥degv, and hencedegu= degv=k. Note that so far we have not made use of the absence of a politician.
Step 2 - All vertices have equal degree. Consider all neighborsw1,· · · , wk of ufrom step one. Since all exceptw2 are non-adjacent tov, we know by step 1 that all exceptw2 have degreek.
Now, since w2 is not has a politician, it has a non-neighborx. Thisxmust be non-adjacent to either u or v by the friendship condition, so by step 1 we have degw2 = degx= k. We thus have shown that all neighbors of u have degree k, and hencek.
Step 3 - Finding a formula for n. Take any vertex uinG. Since every other vertex has a unique common neighbor with u, we can compute the number of vertices in the graph by counting neighbors of neighbors" of u. By step 2, uand all its neighbors have degreek, so taking the sum of degrees of neighbors ofugives k2. This counts every vertex exactly once - except foruitself, which is counted k times. Accounting for this we get
n=k2−k+ 1.
Looking at this equation, we see that ifk= 1ork= 0thenn= 1and ifk= 2 then n = 3, so by the friendship condition we get a 3-cycle. Since both of these graphs have politician vertices, so we conclude thatk >2.
Remark 2.3. Sincek2≡k (mod 2), this shows thatnis odd. It is also easy to prove thatnmust be odd ifGhas a politician (see below) so already we have shown that any nite graph satisfying the friendship condition has an odd number of vertices.
Step 4 - Eigenvalues of the adjacency matrix. Label all the verticesv1,· · ·, vn ofG, and consider the adjacency matrixA= (aij)dened by
aij =
(1 vi adjacent tovj
0 otherwise.
This matrix is obviously symmetric, with 0 on the diagonal (no vertices are adjacent to themselves), and precisely k 1s in each row and column (since the degree of every vertex is k). Also, by the unique common neighbor condition, in any two rows there is exactly one column where both have a 1. Hence
A2=
k 1 · · · 1 1 k · · · 1 ... ... ...
1 1 · · · k
= (k−1)In+J
whereIn is then×nidentity matrix andJ is then×nmatrix consisting entirely of1s.
It is not hard to guess the eigenvectors ofJ. We have
1 1 · · · 1 1 1 · · · 1 ... ... ...
1 1 · · · 1
·
−1
−1...
n−1
= 0
and permuting the entries of this vector we quickly getn−1 linearly independent eigenvectors ofAwith eigenvalue0. Also
1 1 · · · 1 1 1 · · · 1 ... ... ...
1 1 · · · 1
·
1 1...
1
=n·
1 1...
1
so J also has an eigenvaluenof multiplicity 1. It immediately follows thatAhas eigenvaluesk−1of multiplicityn−1 andn+k−1of multiplicity1. Recall from step 3 thatn+k−1 =k2.
NowAis symmetric, so it is diagonalizable to a matrix whose (diagonal) entries are the eigenvalues ofA. The same is true ofA2, and therefore the eigenvalues of A2 are squares of the eigenvalues ofA. HenceAhas eigenvalues±kof multiplicity 1 (multiplyingA with a vector of 1s we see that the eigenvalue is in fact k), and eigenvalues √
k−1 of multiplicity r and −√
k−1 of multiplicity s, for some r, s withr+s=n−1.
Step 5. Since the trace ofAis 0, and the trace is the sum of the eigenvalues, we have
k+ (r−s)√
k−1 = 0.
Sincek >2 we concluder6=sand hence
√
k−1 = k s−r so√
k−1∈Nby Dedekind's theorem. Settingh=√
k−1 we get k=h2+ 1
and re-writing our equation for the trace ofA, k=h(s−r).
sohdividesh2+ 1. Hence h= 1 and thereforek= 2, which we have proved to be impossible.
3. The emergence of windmills
Once we have exposed the politician, it is not hard to see what form the graph must take. Letprepresent the politician vertex, andube any other vertex in the graph. pand u must have a unique common neighbor v. Ifv is adjacent to any other vertexxthen pis also adjacent to x, since pis a politician, and so v andp will have two common neighborsuandx, a contradiction.
Therefore the only neighbors ofv arepand u, and likewise the only neighbors ofuarepandv. We end up with a windmill" similar to gure 1
Since every vertex u6=p can be paired o with another vertexu1 in this way, the graph must have an odd number of vertices.
4. Extension of the problem
Denition 4.1 (Path). We dene a path inGto be a sequence of (not necessarily distinct) verticesv1, v2,· · · in Gsuch that everyvi is adjacent tovi+1 for everyi. We dene the length of the path to be the length of the sequence.
It is clear that our friendship condition is equivalent to saying that there is a unique path of length l = 2 between any two vertices in G. We have shown that the only possible graph satisfying this condition is a windmill.
It has been conjectured by Anton Kotzig that there are no nite graphs which satisfy this condition for anyl >2. Kotzig's conjecture has been veried forl≤33, but the situation for arbitraryl remains something of a mystery.
References
[1] Martin Aigner and Günter M. Ziegler. Proofs from The Book. Springer-Verlag, Berlin, fth edition, 2014. Including illustrations by Karl H. Hofmann.
